Question: Factor completely. $49x^2-9=$
Solution: Both $49x^2$ and $9$ are perfect squares, since $49x^2=({7x})^2$ and $9=({3})^2$. $49x^2-9 = ({7x})^2-({3})^2$ So we can use the difference of squares pattern to factor. ${a}^2 - {b}^2 =({a}+{b})({a}-{b})$ In this case, ${a}={7x}$ and ${b}={3}$ : $({7x})^2 - ({3})^2 =({7x}+{3})({7x}-{3})$ In conclusion, $49x^2-9=(7x+3)(7x-3)$ Remember that you can always check your factorization by expanding it.